Question

The magnetic field due to the earth has a horizontal component of 26 μT at a place where the dip is 60°. Find the vertical component and the magnitude of the field.

Answer 1

The vertical component and the magnitude of the field Bv  = 45  μT

Explanation:

BH = 26μT   = 26×10−6 T

We know that

Here s(dip)=60°

Since BH = Bcos(60°)

26×10−6 T = B x cos(60°)

B =   26×10−6 / 1/2

B =   52 ×10−6 T

B =   52 μT

For vertical component:

Bv = Bsinθ

Bv =  52 ×10−6 sin(60°)

Bv = 44.98 μT

Bv  = 45  μT

The vertical component and the magnitude of the field Bv  = 45  μT

Learn more

What is magnetic field?

https://brainly.in/question/6857847

Answer 2

The vertical component of the Earth's magnetic field By is $45 \mu T$

Explanation:

Step 1:

Given data in the question

Horizontal part of a magnetic field on Earth, $B_{H}=B \cos \delta$

Dip Angle, δ = 60°

Step 1:

Earth's horizontal magnetic field $B_H$ is given by

$B_{H}=B \cos \delta$

B = net earth magnetic field  

When the respective values are replaced, we get

           $26 \times 10^{-6}=B \times \frac{1}{2}$

    $B=52 \times 10^{-6}=52 \mu T$

Step 2:

The vertical component of the Earth's magnetic field $B_y$ is given by

          $B_{y}=B \sin \delta$

          $B_{y}=52 \times 10^{-6} \times \frac{\sqrt{3}}{2}$

          $B_{y}=44.98 \mu T$

          $B_{y}=45 \mu T$